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I am trying to obtain the two tangent lines to an ellipse passing through a certain point. I found some similar questions but none that I could find used the ellipse equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F$.

Looking at the below example, I want to obtain the two points E and F where the ellipse and the point D are known. How am I able to calculate those tangent lines (or points) for a rotated ellipse?

enter image description here

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Translate the plane so that $D$ comes to the origin. The conic equation becomes

$$A(x-x_D)^2+B(x-x_D)(y-y_D)+C(y-y_D)^2+D(x-x_D)+E(y-y_D)+F=0$$

and you can compute the new coefficients.

Now assume that the equation of the tangent is

$$y=mx$$ and you get the condition

$$(cm^2+bm+a)x^2+(em+d)x+f=0.$$

We have tangency when this equation has a double root, i.e.

$$(em+d)^2-4f(cm^2+bm+a)=0.$$

This gives us the solutions

$$m=\frac{\pm\sqrt{(e^2-4cf) (4af-d^2)+(de-2 bf)^2}+2bf-de}{e^2-4cf}.$$

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  • $\begingroup$ Ended up using this. Thanks! $\endgroup$ – Stan Jul 9 at 9:20
  • $\begingroup$ @Stan: glad to know. Not my favorite approach though, as it is not isotropic (and fails for vertical tangents). $\endgroup$ – Yves Daoust Jul 9 at 10:27
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  1. Purely algebraic approach: from equation of ellipse $f(x,y)=0$ and equation of line $(y-y_0)=\alpha(x-x_0)$, you can get an quadratic equation: $$ \frac{y-y_0}{x-x_0}=-\frac{f'_x}{f'_y} $$ which is the equation of ellipse (which is the same as given one, but scaled and shifted). From this, you can find its intersection with the given ellipse.

  2. A better approach. You find canonical parameters of ellipse and thus the affine transformation to transform it to unit circle. You apply this transformation to point $D\to D'$ and find its tangent points $E'$ and $F'$ onto unit circle with simple geometry. Inverse transformation will give you points $E$ and $F$, since affine transformation doesn't change the tangency property.

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Joachimsthal's Notations

  • $s_{ij}=Ax_i x_j+B\left( \dfrac{x_i y_j+x_j y_i}{2} \right)+C y_i y_j+D\left( \dfrac{x_i+x_j}{2} \right)+E\left( \dfrac{y_i+y_j}{2} \right)+F$

  • $s_{i}=Ax_i x+B\left( \dfrac{y_i x+x_i y}{2} \right)+C y_i y+D\left( \dfrac{x+x_i}{2} \right)+E\left( \dfrac{y+y_i}{2} \right)+F$

  • $s=Ax^2+Bxy+C y^2+Dx+Ey+F$

Tangent pairs through $P(x_1,y_1)$:

$$s_1^2=s_{11} s$$

See the proofs here and another answer of mine here.

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